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Svetradugi [14.3K]
3 years ago
6

A hot-water radiator has a surface temperatue of 80 o C and a surface area of 2 m2 . Treating it as a blackbody, find the net ra

te at which it will radiate energy to a 20 o C room (see problem in 1.36)
Physics
1 answer:
faust18 [17]3 years ago
3 0

Answer:

925.04 J/s

Explanation:

T = 80 C = 80 + 273 = 353 K

To = 20 c = 20 + 273 = 293 K

A = 2 m^2

Use the formula for Stefan's law

Energy radiated per second

E = \sigma  A \left ( T^{4}-T_{0}^{4} \right )

E = 5.67 \times 10^{-8}\times 2\left ( 353^{4}-293^{4} \right )

E = 925.04 J/s

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work done as mass 1 moves to mass 2. the gravitational force between two point masses separated by a distance r is proportional
pav-90 [236]

Answer:

gravitational force

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3 years ago
A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 m from the bulb, the light in
krek1111 [17]
1. If we increase the distance to twice it's original value, the light intensity is reduced by one-fourth, the light intensity would be:
I0/4

2. rms magnetic field is inversely proportional to distance, so the new rms magnetic field would be:
B0/2

3. average energy density is inversely proportional to the square of the distance, so the new average energy density is:
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5 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
sdas [7]

Answer:

0.34 m

Explanation:

From the question,

v = λf................ Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f............... Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But,  the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

6 0
3 years ago
When a cracker or bread dissolves in your mouth, is that a physical or chemical change?
BabaBlast [244]
Physical because ur making dissolve
8 0
3 years ago
Read 2 more answers
Capacitors C1 = 6.45 µF and C2 = 2.50 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon
denpristay [2]

Answer:

For C1, Q =  1.6125×10⁻³ C

For C2, Q =  6.25×10⁻⁴ C

Explanation:

Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.

From the question,

Q = CV........................ Equation 1

Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.

For the first capacitor,

Q = C1V............. Equation 2

Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V

Substitute into equation 2

Q = (6.45×10⁻⁶ )(250)

Q = 1.6125×10⁻³ C.

For the the second capacitor,

Q = C2V............. Equation 3

Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V

Q = (2.5×10⁻⁶ )(250)

Q = 6.25×10⁻⁴ C

4 0
3 years ago
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